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Write electron configurations to show the first two ionization processes for potassium. Explain why the second ionization energy is much greater than the first.

Short Answer

Expert verified
The second ionization energy of potassium is much greater than the first because it involves removing an electron from a stable, full shell.

Step by step solution

01

Write Electron Configuration for Neutral Potassium

The atomic number of potassium (K) is 19. Therefore, the electron configuration for neutral potassium is \[ 1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 \]. This means that it has one electron in the 4s sublevel, which is relatively easy to remove.
02

First Ionization Process of Potassium

In the first ionization process, potassium loses one electron from the 4s orbital to form a \( K^+ \) ion. The electron configuration for \( K^+ \) is \[ 1s^2 2s^2 2p^6 3s^2 3p^6 \]. This is because the electron removed comes from the highest energy level, which is the 4s sublevel with one electron.
03

Second Ionization Process of Potassium

For the second ionization process, we attempt to remove another electron. The \( K^+ \) ion already has a stable noble gas configuration similar to argon: \[ 1s^2 2s^2 2p^6 3s^2 3p^6 \]. Removing another electron would mean taking an electron from this full and stable configuration, generally from the 3p sublevel.
04

Explain the Difference in Ionization Energies

The second ionization energy is much greater than the first because the first ionization involves removing an electron from the 4s orbital, which is a single electron in a higher energy, less stable state. The second ionization energy involves removing an electron from a complete and stable electron shell configuration (similar to argon), which requires significantly more energy.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ionization Energy
Ionization energy refers to the amount of energy required to remove an electron from an atom or ion in its gaseous state. It is an important concept when discussing electron configurations and periodic table trends. Upon ionization, an electron is removed from an atom, which requires energy.
The first ionization energy is the energy needed to remove the first electron. This first electron is generally from the outermost shell where electrons are most free and least tightly held by the nucleus.
For potassium, with a neutral electron configuration of \[ 1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 \], the first ionization energy involves removing the single electron in the 4s orbital.
  • This 4s electron is further from the nucleus and shielded by inner shell electrons, making it relatively easy to remove.
The second ionization energy is substantially larger because it requires removing an electron from a filled and stable shell. When energy is needed to remove an electron from a stable electron configuration, such as the noble gas configuration of argon, more energy is required because the electrons are more tightly bound to the nucleus.
Potassium Ions
Potassium ions play a significant role in chemical reactions and biological systems. When discussing ionization, potassium usually forms a \( K^+ \) ion.
This involves removing one electron from the outermost electron shell. After this first electron is removed, the \( K^+ \) ion has the electron configuration of \[ 1s^2 2s^2 2p^6 3s^2 3p^6 \],
making it isoelectronic with the noble gas argon. This is a stable configuration as it fills all the available orbitals in the energy level.
  • In forming ions, elements like potassium aim to achieve a stable electron configuration, typically resembling that of a noble gas.
  • The formation of a stable \( K^+ \) ion is why potassium tends to prefer a +1 charge in its compounds.
Attempting to remove one more electron to form \( K^{2+} \) would disturb this noble gas configuration,
requiring a substantial increase in energy.
Noble Gas Configuration
A noble gas configuration is achieved when an atom has a full outer electron shell, making it exceptionally stable. In the periodic table, noble gases such as helium, neon, and argon, have full s and p orbitals in their outer electron shells.
Potassium, when neutral, has an electron configuration of \[ 1s^2 2s^2 2p^6 3s^2 3p^6 4s^1 \].
Upon the first ionization, which involves the removal of the 4s electron, potassium becomes \( K^+ \) with an electron configuration similar to argon, \[ 1s^2 2s^2 2p^6 3s^2 3p^6 \], and achieves a noble gas configuration.
  • This is why the removal of a second electron is energetically unfavorable, as it disrupts this stable structure.
  • Electrons in a noble gas configuration are particularly stable due to the complete octet in the outermost shell.
Achieving a noble gas configuration is a driving factor in the chemical behavior of elements, often guiding the formation of ions and bonds in compounds.

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Most popular questions from this chapter

The discovery of two new elements (atomic numbers 113 and 115 ) was announced in February 2004 (IMAGE NOT COPY) (a) Use spdfand noble gas notations to give the electron configurations of these two elements. (b) Name an element in the same periodic group as the two elements. (c) Element 113 was made by firing a light atom at a heavy americium atom. The two combine to give a nucleus with 113 protons. What light atom was used as a projectile?

Two elements in the second transition series ( \(Y\) through Cd) have four unpaired electrons in their \(3+\) ions. What elements fit this description?

Compare the elements Na, B, Al, and C with regard to the following properties: (a) Which has the largest atomic radius? (b) Which has the most negative electron affinity? (c) Place the elements in order of increasing ionization energy.

Which of the following is not an allowable set of quantum numbers? Explain your answer briefly. \(n \quad \ell \quad m_{\ell} \quad m_{\mathrm{s}}\) \(\begin{array}{lllll}\text { (a) } & 2 & 0 & 0 & -\frac{1}{2}\end{array}\) \(\begin{array}{lllll}\text { (b) } & 1 & 1 & 0 & +\frac{1}{2}\end{array}\) \(\begin{array}{lllll}\text { (c) } & 2 & 1 & -1 & -\frac{1}{2}\end{array}\) \(\begin{array}{lllll}\text { (d) } 4 & 3 & +2 & -\frac{1}{2}\end{array}\)

Explain briefly why each of the following is not a possible set of quantum numbers for an electron in an atom. In each case, change the incorrect value (or values) to make the set valid. (a) \(n=4, \ell=2, m_{\ell}=0, m_{\mathrm{s}}=0\) (b) \(n=3, \ell=1, m_{\ell}=-3, m_{s}=-\frac{1}{2}\) (c) \(n=3, \ell=3, m_{\ell}=-1, m_{\mathrm{s}}=+\frac{1}{2}\)

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